Integrate. $ \int -9\cos(x)\,dx $ $=$ $+ C$
Explanation: We need a function whose derivative is $-9\cos(x)$. We know that the derivative of $\sin(x)$ is $\cos(x)$, so let's start there: $\dfrac{d}{dx} \sin(x) = \cos(x)$ Now let's add in a negative sign: $\dfrac{d}{dx}\left[ -\sin(x)\right] = -\cos(x)$ Now let's multiply by $9$ : $\dfrac{d}{dx}\left[ -9\sin(x) \right]= -9\cos(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int -9\cos(x)\,dx =-9 \sin(x)\, + C$ The answer: $-9\sin(x)+C$.